Integrand size = 19, antiderivative size = 40 \[ \int \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*(2*A+C)*arctanh(sin(d*x+c))/d+1/2*C*sec(d*x+c)*tan(d*x+c)/d
Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A \coth ^{-1}(\sin (c+d x))}{d}+\frac {C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 d} \] Input:
Integrate[Sec[c + d*x]*(A + C*Sec[c + d*x]^2),x]
Output:
(A*ArcCoth[Sin[c + d*x]])/d + (C*ArcTanh[Sin[c + d*x]])/(2*d) + (C*Sec[c + d*x]*Tan[c + d*x])/(2*d)
Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 4534, 3042, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {1}{2} (2 A+C) \int \sec (c+d x)dx+\frac {C \tan (c+d x) \sec (c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (2 A+C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {C \tan (c+d x) \sec (c+d x)}{2 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {(2 A+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {C \tan (c+d x) \sec (c+d x)}{2 d}\) |
Input:
Int[Sec[c + d*x]*(A + C*Sec[c + d*x]^2),x]
Output:
((2*A + C)*ArcTanh[Sin[c + d*x]])/(2*d) + (C*Sec[c + d*x]*Tan[c + d*x])/(2 *d)
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.38
method | result | size |
derivativedivides | \(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(55\) |
default | \(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(55\) |
parts | \(\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(57\) |
parallelrisch | \(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+C \sin \left (d x +c \right )}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(86\) |
norman | \(\frac {\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {\left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 A +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(93\) |
risch | \(-\frac {i C \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\) | \(118\) |
Input:
int(sec(d*x+c)*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(A*ln(sec(d*x+c)+tan(d*x+c))+C*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d *x+c)+tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.80 \[ \int \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, C \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
Output:
1/4*((2*A + C)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*A + C)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*C*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(A+C*sec(d*x+c)**2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*sec(c + d*x), x)
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.45 \[ \int \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, A + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, C \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \] Input:
integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
Output:
1/4*((2*A + C)*log(sin(d*x + c) + 1) - (2*A + C)*log(sin(d*x + c) - 1) - 2 *C*sin(d*x + c)/(sin(d*x + c)^2 - 1))/d
Time = 0.34 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.50 \[ \int \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, A + C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (2 \, A + C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, C \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \] Input:
integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2),x, algorithm="giac")
Output:
1/4*((2*A + C)*log(abs(sin(d*x + c) + 1)) - (2*A + C)*log(abs(sin(d*x + c) - 1)) - 2*C*sin(d*x + c)/(sin(d*x + c)^2 - 1))/d
Time = 11.32 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.02 \[ \int \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A+\frac {C}{2}\right )}{d}-\frac {C\,\sin \left (c+d\,x\right )}{2\,d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \] Input:
int((A + C/cos(c + d*x)^2)/cos(c + d*x),x)
Output:
(atanh(sin(c + d*x))*(A + C/2))/d - (C*sin(c + d*x))/(2*d*(sin(c + d*x)^2 - 1))
Time = 0.21 (sec) , antiderivative size = 177, normalized size of antiderivative = 4.42 \[ \int \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -\sin \left (d x +c \right ) c}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)*(A+C*sec(d*x+c)^2),x)
Output:
( - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c + 2*log(tan((c + d*x)/2) - 1)*a + log(tan((c + d*x)/ 2) - 1)*c + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + log(tan((c + d *x)/2) + 1)*sin(c + d*x)**2*c - 2*log(tan((c + d*x)/2) + 1)*a - log(tan((c + d*x)/2) + 1)*c - sin(c + d*x)*c)/(2*d*(sin(c + d*x)**2 - 1))